Find the area?

Question

 

Find the area of the region enclosed by the polar curve r=1-cos θ.

 

Answer: (3/2)pi

Answer 1

Area

= ∫{0, 2pi} (1/2)r² dθ

= ∫{0, 2pi} (1/2)(1-cosθ)² dθ

= ∫{0, 2pi} (1/2)(1-2cosθ+cos²θ) dθ

= ∫{0, 2pi} (1/2)(1-2cosθ+.5(1+cos2θ)) dθ

= ∫{0, 2pi} (1/2)(1.5) dθ

= (3/2)pi <==Answer

 

Answer 2

Reading Robert's answer, I had to figure out where he got some of his steps.

 

Think of two rays separated by a very small angle, dθ, intersecting the curve. The area enclosed by the rays and curve can be approximated by a circular sector with radius r and arc length r*dθ. The area of that little sector is a fraction of a whole circle,
(π*r^2)*(r*dθ)/(2πr) = (1/2)*(r^2*dθ).

Adding them up over one rotation:
Area = ∫{0,2π} (1/2)*r^2 dθ
= (1/2) ∫{0,2π} (1 - cos(θ))^2 dθ
= (1/2) ∫{0,2π} (1 - 2cos(θ) + cos^2(θ)) dθ

cos(2θ) = 2 cos^2(θ) - 1
cos^2(θ) = (1/2)*(1 + cos(2θ))

Area = (1/2) ∫{0,2π} (1 - 2cos(θ) + (1/2)*(1 + cos(2θ))) dθ
= (1/2) ∫{0,2π} (3/2 - 2cos(θ) + (1/2)*cos(2θ) ) dθ
= (1/2) [(3/2)*θ - 2sin(θ) + (1/4)*sin(2θ)]{0,2π}
= (1/2) [(3/2)*2π - 2sin(2π) + (1/4)*sin(4π) - (3/2)*0 + 2sin(0) - (1/4)*sin(0)]
= (1/2) [3π]
= 3π/2